How can I calculate the compound angle when I know two of the angles but don't know what the angle would be where the two intersect?
If "A" and "C" are sitting on a 9.16 degree angle and the are coming together at a 90 degree angle in the corner, how can the compound angle of "A" and "C" be determined? In the past, I would try 3d modeling such a space and testing the angles and trying to figure it out by stepping backwards, but I feel that there must be a simpler and more elegant solution.
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If we choose axis $x$ along the separation line between $A$ and $B$, axis $y$ along the separation line between $C$ and $D$, axis $z$ perpendicular to plane $BD$, then it is not difficult to find two unit vectors normal to $A$ and $C$, both directed "inwards": $$ n_A=(0,\sin\theta,-\cos\theta), \quad n_C=(\sin\theta,0,-\cos\theta), $$ where $\theta$ is the dihedral angle between planes $A/B$ (and $C/D$).
The angle $\alpha$ between the vectors satisfies then $\cos\alpha=n_A\cdot n_C=\cos^2\theta$. The dihedral angle $\phi$ between planes $A/C$ is the supplementary of $\alpha$ and we have then the very simple relation: $$ \cos\phi=-\cos^2\theta. $$ For the given example: $$ \cos\phi=-\cos^2 99.16°=-0.0253421, \quad \phi=91.45°. $$
One method would be to construct vectors $n_A$ and $n_C$ perpendicular to the two surfaces $A$ and $C,$ then use the formula $$ n_A \cdot n_C = \|n_A\| \|n_C\| \cos \theta_{AC}. $$ Solve this for $\theta_{AC},$ the angle between the surfaces.
This formula works reasonably well in the given example since the angle between the surfaces is not too far from $90$ degrees. Angles near $90$ degrees give good results because $\frac{d}{d\theta} \cos\theta$ is maximized when $\theta$ is $90$ degrees. If you try this method for surfaces that are nearly parallel, you run into the fact that when $\cos\theta \approx 1,$ a small error in the value of $\cos\theta$ translates into a much larger error in the value of $\theta.$